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The number of moles of NaCl in 3 litres ...

The number of moles of NaCl in 3 litres of 3 M solution is:

A

1

B

3

C

9

D

27

Text Solution

Verified by Experts

The correct Answer is:
C
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The number of mole of solute per litre solution is called

Knowledge Check

  • When the concentration is expressed as the number of moles of a solute per litre of solution it is known as

    A
    Normality
    B
    Molarity
    C
    Mole fraction
    D
    Mass percentage

  • The number of moles of KCI in 1000 mL of 3M solution is ______.

    A
    1
    B
    2
    C
    3
    D
    `1.5`

  • The number of moles of NaCl in 250cm^(3) of 0.50 M NaCl is

    A
    0.250 mol
    B
    2 mol
    C
    0.125 mol
    D
    1 mol

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    Number of moles of solute dissolved per litre of solution is

    Assertion : When a concentrated solution is diluted by adding more water , the number of moles of solute in the solution remains unchanged . Reason : Number of a solute is equal to the product of molarity and volume of solution in litres.

    The number of moles present in 2 litre of 0.5 M NaOH is:

    The number of millimoles present per litre of solution of CH_3COONa so that 150 ppm (150 moles of its gets hydrolysed out of one million moles added) of if K_a=1.8xx10^(-5) is :

    The concentrations of soluitons can be expressed in number of ways , viz : mass fraction of solute (or mass percent), Molar concentration (Molarity ) and Molal concentration (molality). These terms are known as concentration terms and also they are related with each otehr i.e., knowing one concentration terms for the solution, we can find other concentration terms also. the definition of different cencentration terms are given below: Molarity : It is number of moles of solute present in one litre of the solution. Molality : It is the number of moles of solute present in one kg of the solvent. Mole fraction =("Mole of solute")/("Moles of solute" + "Moles of solvent") If molality of the solution is given as a, then mole fraction of the solute can be calculated by Mole Fraction =(a)/(a+(100)/(M_("solvent"))),=(axxM_("solvent"))/((a xx M_("solvent")+1000)0 where a=molality and M_("solvent") =Molar mass of solvent We can change : Mole fraction hArr Molality hArr Molarity Percentage (weight/vol) of NaCl persent in the solution is :

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